Simple Proof of Fermat's Last Theorem and Beal's Conjecture

MrAwojobi
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Attached is the proof.
 

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I forgot to mention that n is odd and A and B are not necessarily integers in the transformation. This slight ammendments are in this new attachment.
 

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MrAwojobi said:
A simple argument shows why A^n + B^n ≠ (P+Q)^n if n is odd.

Please provide this argument. I mean, I can give proofs like this to. You simply leave out the hard part...


Obviously, if n is even, A^n + B^n = C^n collapses to the Pythagorean equation.

I do not see this. Please provide a proof...
 
I have revamped the proof again so that n is not required to be odd.
 

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MrAwojobi said:
I have revamped the proof again so that n is not required to be odd.

The statement "it should be clear that since the coefficients referred to previously are set in a particular way for each part, the only way they can equal each other is if C and B and therefore A have a common prime factor" is not proven. It is merely a restatement of Beal's conjecture.
 
ramsey2879

I have added an extra paragraph to my article and this offers more explanations to my claims.
 

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Thread 'Trouble understanding an online solution to an exercise in Dummit & Foote'

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The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
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Thread 'Decomposition into irreps of compact Lie group'

When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
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